Answer:
The correct answer is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.
Explanation:
1. Let's check all the information given to us to answer the question correctly:
Mean of the scores on the aptitude test = 100
Standard deviation of the aptitude test = 4.6
Felix's score on the aptitude test = 106
Mean of the scores on the knowledge test = 70
Standard deviation of the knowledge test = 2.4
Felix's score on the knowledge test = 75
2. Which statement best describes Felix's scores on the two tests comparatively?
Let's recall that z-score in a normal distribution, positive or negative, is the number of times of the standard deviation a certain element is from the mean. If the element is below the mean, then the z-score is negative and if it's above the mean, then the z-score is positive.
Therefore, a score of 106 on the aptitude test, will have the following z-score:
106 - 100 = 6 and it's above the mean.
Now, we calculate the z-score, using the value of the standard deviation this way:
6/4.6 = 1.30
A score of 75 on the knowledge test, will have the following z-score:
75 - 70 = 5 and it's above the mean.
Now, we calculate the z-score, using the value of the standard deviation this way:
5/2.4 = 2.08
The z-scores of Felix were 1.30 on the aptitude test and 2.08 on the knowledge test. He performed better on the aptitude test because 2.08 > 1.30, so the correct statement that best describes Felix's scores on the two tests comparatively is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.